高数:图片上的不定积分咋求(过程详细些)?
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(sint)^6 (tant)^2 = (1-(cost)^2)(sint)^4 (tant)^2 = (sint)^4 (tant)^2 -(sint)^6
=(sint)^2 (tant)^2 - (sint)^4 - (sint)^6
= (tant)^2 - (sint)^2 - (sint)^4 - (sint)^6 = (sect)^2 -1 - (sint)^2 - (sint)^4 - (sint)^6
∫(sint)^6 (tant)^2dt = ∫(sect)^2 -1 - (sint)^2 - (sint)^4 - (sint)^6
∫(sect)^2 dt = (tant)^2 +C
∫(sint)^2 dt = 0.5 ∫ 1-cos2t dt = 0.5 t - 0.25 ∫ cos2t d2t =0.5t -0.25 sin2t +C
∫(sint)^4 dt = ∫(1-cos2t)^2/4 dt =0.25∫1 -2cos2t + (cos4t)^2 dt
= 0.25 t -0.25 ∫ cos2t d2t +0.125 ∫ 1+cos4t dt
=0.25 t - 0.25 sin2t +0.125 t +1/32 sin4t +C
= 0.375t -0.25 sin2t + 1/32 sin4t +C
∫(sint)^6 dt = 1/8 ∫ (1-cos2t)^3 dt = 1/8 ∫1 -3cos2t +3(cos2t)^2 -(cos2t)^3dt
然后利用类似的方法继续求
=(sint)^2 (tant)^2 - (sint)^4 - (sint)^6
= (tant)^2 - (sint)^2 - (sint)^4 - (sint)^6 = (sect)^2 -1 - (sint)^2 - (sint)^4 - (sint)^6
∫(sint)^6 (tant)^2dt = ∫(sect)^2 -1 - (sint)^2 - (sint)^4 - (sint)^6
∫(sect)^2 dt = (tant)^2 +C
∫(sint)^2 dt = 0.5 ∫ 1-cos2t dt = 0.5 t - 0.25 ∫ cos2t d2t =0.5t -0.25 sin2t +C
∫(sint)^4 dt = ∫(1-cos2t)^2/4 dt =0.25∫1 -2cos2t + (cos4t)^2 dt
= 0.25 t -0.25 ∫ cos2t d2t +0.125 ∫ 1+cos4t dt
=0.25 t - 0.25 sin2t +0.125 t +1/32 sin4t +C
= 0.375t -0.25 sin2t + 1/32 sin4t +C
∫(sint)^6 dt = 1/8 ∫ (1-cos2t)^3 dt = 1/8 ∫1 -3cos2t +3(cos2t)^2 -(cos2t)^3dt
然后利用类似的方法继续求
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